Subject: Re: [xsl] XPath MOD 10 calculation From: Abel Braaksma <abel.online@xxxxxxxxx> Date: Thu, 24 May 2007 13:46:18 +0200 |
Well, 'elegant' is in the eye of the beholder, I think ;)
Perhaps this is something that you consider more elegant. But depending on your audience, it may win the beauty contest, but not receive praise for elegance:
<xsl:variable name="x"
select="sum(for $i in (1,3,5,7,9,11,2,4,6,8,10) return number(substring(., $i, 1))) mod 10" />
and leave the xsl:if as-is. Or you can rewrite your xsl:if like this:
<xsl:if test="(10 - $x) mod 10 ne substring(., 12, 1)" />
which brings us to a possibility to combine the two in one expression without repetition:
<xsl:if test="(10 - sum(for $i in (1,3,5,7,9,11,2,4,6,8,10) return number(substring(., $i, 1))) mod 10) mod 10 ne substring(., 12, 1)" />
I'm not certain of the order of operator precedence, you may have to insert some parentheses here and there. I didn't test it, because I don't know the MOD 10 algorithm and you did not provide enough examples to really test it. I chose for number() instead of xs:integer because sum() takes doubles and xs:integer can fail with an error, where number() cannot fail.
Cheers, -- Abel Braaksma
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